Q. From a circular disc of radius R and mass 9M, a small disc of mass M and radius R/3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is
A) (40/9) MR²
B) MR²
C) 4 MR²
D) (4/9) MR²
Explanation
[Solution from source] Full disc: I₁ = (9M)R²/2. Removed disc: I₂ = M(R/3)²/2 = MR²/18. Remaining I = 81MR²/18 − MR²/18 = 80MR²/18 = (40/9)MR². Answers like 4MR² come from subtracting MR²/2 instead of using the small disc's own radius R/3. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.